Factoring Special Cases ALGEBRA 1 LESSON 9-7 (For help, go to Lessons 8–4 and 9-4.) Simplify each expression. 1. (3x)2 2. (5y)2 3. (15h2)2 4. (2ab2)2 Simplify each product. 5. (c – 6)(c + 6) 6. (p – 11)(p – 11) 7. (4d + 7)(4d + 7) 5-5

Factoring Special Cases ALGEBRA 1 LESSON 9-7 Solutions 1. (3x)2 = 32 • x2 = 9x2 2. (5y)2 = 52 • y2 = 25y2 3. (15h2)2 = 152 • (h2)2 = 225h4 4. (2ab2)2 = 22 • a2 • (b2)2 = 4a2b4 5. (c – 6)(c + 6) is the difference of squares. (c – 6)(c + 6) = c2 – 62 = c2 – 36 6. (p – 11)(p – 11) is the square of a binomial. (p – 11)2 = p2 – 2p(11) + 112 = p2 – 22p + 121 7. (4d + 7)(4d + 7) is the square of a binomial. (4d + 7)2 = (4d)2 + 2(4d)(7) + 72 = 16d2 + 56d + 49 5-5

Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor a2 – 16. a2 – 16 = a2 – 42 Rewrite 16 as 42. = (a + 4)(a – 4) Factor. Check: by multiplication. (a + 4)(a – 4) a2 – 4a + 4a – 16 a2 – 16 5-5

Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor 9b2 – 25. 9b2 – 225 = (3b)2 – 52 Rewrite 9b2 as (3b)2 and 25 as 52. = (3b + 5)(3b – 5) Factor. 5-5

Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor 5x2 – 80. 5x2 – 80 = 5(x2 – 16) Factor out the GCF of 5. = 5(x + 4)(x – 4) Factor (x2 – 16). Check: Multiply the binomials. Then multiply by the GCF. 5(x + 4)(x – 4) 5(x2 – 16) 5x2 – 80 5-5

Factoring Special Cases ALGEBRA 1 LESSON 9-7 Factor each expression. 1. y2 – 18y + 81 2. 9a2 – 24a + 16 3. p2 – 169 4. 36x2 – 225 5. 5m2 – 45 6. 2c2 + 20c + 50 (y – 9)2 (3a – 4)2 (p + 13)(p – 13) (6x + 15)(6x – 15) 5(m + 3)(m – 3) 2(c + 5)2 5-5